Video: Finding the Integration of a Function Containing Exponential Functions by Distributing the Division

Determine ∫ (8𝑒^(3π‘₯) βˆ’ 𝑒^(2π‘₯) + 9)/(7𝑒^(π‘₯)) dπ‘₯.

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Video Transcript

Determine the indefinite integral of eight 𝑒 to the three π‘₯ minus 𝑒 to the two π‘₯ plus nine over seven 𝑒 to the power of π‘₯ with respect to π‘₯.

Now, this question might look really nasty, and you might be starting to consider how a substitution might help. However, it’s important to notice that we can simplify the integrand by simply dividing each part of the numerator by seven 𝑒 to the power of π‘₯, remembering that we can then simply subtract the exponents. When we do, we’re left with the indefinite integral of eight-sevenths 𝑒 to the two π‘₯ minus one-seventh 𝑒 to the π‘₯ plus nine-sevenths 𝑒 to the negative π‘₯.

Next, we recall that we can separate this integral. The integral of the sum of a number of functions is equal to the sum of the integrals of each function. And we can also take any constant factors outside of the integral and deal with them later. So, we’re looking at eight-sevenths of the integral of 𝑒 to the two π‘₯ with respect to π‘₯ minus one-seventh of the integral of 𝑒 to the π‘₯ with respect to π‘₯ plus nine-sevenths of the integral of 𝑒 to the power of negative π‘₯ with respect to π‘₯. Okay, So what next?

Well, we know that the indefinite integral of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯ plus 𝑐. But what about the integral of 𝑒 of the power of two π‘₯? You might be wanting to make a prediction as to what you think this will give. And there is a standard result that we can quote. But let’s look at this derivation using integration by substitution. We’re going to let 𝑒 be equal to two π‘₯. So, d𝑒 by dπ‘₯ is equal to two. Now, we know that d𝑒 by dπ‘₯ is absolutely not a fraction, but we can treat it a little like one for the purposes of integration by substitution. And we see that a half d𝑒 is equal to dπ‘₯.

And then, we see that we can simply replace two π‘₯ with 𝑒 and dπ‘₯ with a half d𝑒 and then take out this factor of one-half. And all we need to do now is integrate 𝑒 to the power of 𝑒 with respect to 𝑒. Well, that’s simply a half times 𝑒 to the power of 𝑒. But of course, since 𝑒 is equal to two π‘₯, we can say that the indefinite integral of 𝑒 to the power of two π‘₯ is a half times 𝑒 to the power of two π‘₯ plus a constant of integration. Let’s call that 𝐴. And this is great because it provides us with the general result for the integral of 𝑒 the power of π‘Žπ‘₯ for real constants π‘Ž. It’s one over π‘Ž times 𝑒 to the power of π‘Žπ‘₯.

We can use this result. And we see that the integral of 𝑒 to the power of negative π‘₯ is one over negative one times 𝑒 to the power of negative π‘₯ plus 𝑐. So, popping this altogether, we see that our indefinite integral is eight-sevenths times a half 𝑒 to the two π‘₯ plus 𝐴 minus a seventh times 𝑒 to the π‘₯ plus 𝐡 plus nine-sevenths times one over negative one times 𝑒 to the power of negative π‘₯ plus 𝐢. Distributing our parentheses and combining all our constants, and we’re left with the solution as being four-sevenths 𝑒 to the power of two π‘₯ minus 𝑒 to power of π‘₯ over seven minus nine-sevenths 𝑒 to the power of negative π‘₯ plus 𝐷.

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